Whats 3,684,266 to the nearest hundred thousand and million

Which is greater 80 oz or 5 lb

Len [333]

There the same thing they both weigh the same thing hope it helps and have a great day!

3 0

2 years ago

Read 2 more answers

In triangle MNO shown below, segment NP is an altitude:

Effectus [21]

Answer: C. Definition of an Altitude

Step-by-step explanation:

Given: In triangle MNO shown below, segment NP is an altitude from the right angle.

Let ∠MNP=x

Then ∠PNO=90°-x

Therefore in triangle MNO,

∠MPN=∠NPO =90° [by definition of Altitude]

[Definition of altitude : A line which passes through a vertex of a triangle, and joins the opposite side forming right angles. ]

Now using angle sum property in ΔMNP

∠MNP+∠MPN+∠PMN=180°

⇒x+90°+∠PMN=180°

⇒∠PMN=180°-90°-x

⇒∠PMN=90°-x

Now, in ΔMNO and ΔPNO

∠PMN=∠PNO=90°-x

and ∠MPN=∠NPO =90° [by definition of altitude]

Therefore by AA similarity postulate, we have

ΔMNO ≈ ΔPNO

5 0

2 years ago

At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t

grandymaker [24]

Answer:

(a1) The probability that temperature increase will be less than 20°C is 0.667.

(a2) The probability that temperature increase will be between 20°C and 22°C is 0.133.

(b) The probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c) The expected value of the temperature increase is 17.5°C.

Step-by-step explanation:

Let X = temperature increase.

The random variable X follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].

The probability density function of X is:

$f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.$

(a1)

Compute the probability that temperature increase will be less than 20°C as follows:

$P(X$

Thus, the probability that temperature increase will be less than 20°C is 0.667.

(a2)

Compute the probability that temperature increase will be between 20°C and 22°C as follows:

$P(20$

Thus, the probability that temperature increase will be between 20°C and 22°C is 0.133.

(b)

Compute the probability that at any point of time the temperature increase is potentially dangerous as follows:

$P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467$

Thus, the probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c)

Compute the expected value of the uniform random variable X as follows:

$E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5$

Thus, the expected value of the temperature increase is 17.5°C.

7 0

3 years ago

Sedrtfyghujiklmnbvgcfdrt6yuiopl;kmnjbhgvfcdxswertyuioklkmjnhbgvfds

Rama09 [41]

Answer: hshebe rjwkshzbjskxicbxuxbxjdnxkskzbu

Step-by-step explanation:

3 0

2 years ago

Point A to point B: Walk 200 m north for 3 minutes. •

vampirchik [111]

Step-by-step explanation:

200 + 100 + 150 + 200 +150 = 800 m in total

8 min in total

you are going 100 m a minute or 6000 m an hour

6 0

2 years ago